tom wilson, department of geology and geography exponentials and logarithms – additional...

Tom Wilson, Department of Geology and Geography

Exponentials and logarithms – additional

perspectivestom.h.wilson

[email protected]

Department of Geology and GeographyWest Virginia University

Morgantown, WV


In our first lecture we discussed age/thickness relationships. Initially we assumed that the length of time represented by a certain thickness of a rock unit, was a constant for all depths. However, we noted that as a layer of sediment is buried it will be compacted. Water will be squeezed out of pore spaces, the porosity will decrease, and the grains themselves may be deformed through the process of fracturing or pressure solution.

We realized that the age relationship for a certain thickness of rock could not be constant, but must vary in a more complicated way as a function of the depth of burial.

Porosity Depth Relationships


zx 2 6.0

assumes that the initial porosity (0.6) decreases by 1/2 from one kilometer of depth to the next. Thus the porosity () at 1 kilometer is 2-1 or 1/2 that at the surface (i.e. 0.3), (2)=1/2 of (1)=0.15 (I.e. =0.6 x 2-2 or 1/4th of the initial porosity of 0.6.

Equations of the typecxaby

Are referred to as _________ growth laws or exponential functions.

allometric


The porosity-depth relationship is often stated using the natural base e, where e equals 2.71828 ..

In the geologic literature you will often see the porosity depth relationship written as -cz

0 e 0 is the initial porosity, c is a compaction factor and z - the depth.

Sometimes you will see such exponential functions written as -cz

0 exp In both cases, e=exp=2.71828


z

-

0 eWaltham writes the porosity-depth relationship as

Note that since z has units of kilometers (km) that c (in the previous equation) will have units of km -1 and , units of km.

Also recall that when z=,

01-

0

-

0 368.0

ee

Hence, represents the depth at which the porosity drops to 1/e or 0.368 of its initial value.

-cz0 e In the form c is the reciprocal of that depth.


Can you evaluate the natural log of

-cz0 e

? lnln cz-0 e


cz 0ln)ln(

is a straight line.

Power law relationships end up being straight lines when the log of the relationships is taken.

In our next computer lab we’ll determine the coefficients c (or ) and ln(0) that define the straight line relationship above between ln() and z.

We will also estimate power law and general polynomial interrelationships using Excel.


Thus far we’ve worked with four exponential functions

-cz0 e

- t0 a a e

z-

0 Zp p e

t0 N N e

Porosity depth

Radioactive decay

Reef growth rate

Population growth

So exponential functions relatively common


Problem 3.10

1 pg

VV

You’re given that

Where is the bulk density of the rock; g, the grain density, V the total rock volume and Vp is the volume of the pore space.

Question > Can you re-write this expression in terms of the

porosity ?


How would you rewrite this equation if the pore space were filled with water of density w?

1 pg

VV

Next, assume that the porosity varies with depth exponentially as

-cz0 e

Can you develop an exponential representation of the bulk density in terms of & ?


2

2

1

2

1

r

r

v

v

Substitution of known interrelationships to extract information about other variables

Where settling velocity = lake depth/settling time, you were able to show that 2

2

1

1

2

r

r

t

t

2

2 0.1

10 1

t

9

2 2grv

fp


This allowed us to determine that t2

9

2 2grv

fp

If we are given the viscosity of water then we can solve Stokes equation explicitly for the velocity

With a viscosity of water equal to about 0.01 poise [gm/(cm-s)].

In the lab exercise, you are asked to do this for a range of particle sizes

extending from 0.001cm to 0.1cm (0.01mm to 1 mm).


You will get a continuous plot of velocity versus

particle radius.

1 mm radius


1 mm radius

According to Waltham, it took 10 days for the 1mm radius particle to settle to the bottom of the lake. How deep is it?


In the expanded lab version of problem 3.11, you are also asked to develop a plot of settling time versus

particle radius.

9

2 2grv

fp

22

9p f grd

t

29

2 p f

dt r

g

Comment on how the plot of settling time compares to the plot of settling velocity. Think about this in the context of

comments in class about the relationship of Stokes’ equation for velocity compared to the expression modified

to show how time varies with particle radius.


9

2 2grv

fp

29

2 p f

dt r

g

Look at the relationships side-by-side and also consider the logs of v and t?

2v ar 2t br

What kind of relationships do you get?

What do you get when you solve for vt


Complete problems 3.10

and 3.11

Hand in Tuesday

Feb 23rd


Chapter 4 has been on your reading list. We won’t spend much time on the examples in this chapter.

The chapter is short and is a good review of some of the concepts discussed to date.

• In particular - working with exponentials and logarithms


Chapter 4 problems to look over …

In problem 4.7 we have another exponential relationship associated with a geological process – variations in the thickness of the bottom set bed.

x-

0 Xt t e


Chapter 4 problems to look over …

Also look over problem 4.10. Units are always an issue that we shouldn’t loose sight of when we are doing calculations. Using dimensional analysis consider whether the statement

Age =(Depth x Rate) + Age of Top

Has the correct dimensionsThe equation should look familiar. Remember what the slope was in

A=kD+Ao


Let’s return to the lab exercise, but also note

Hand in the take-home isostacy problem before leaving today

Start reviewing class and textbook materials for a test next Thursday

We’ll have a review session next Tuesday

tom wilson, department of geology and geography exponentials and logarithms – additional...

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